Answer:
1.1 × 10² g
Explanation:
First, we will convert 1.0 L to cubic centimeters.
1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³
The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:
1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g
1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:
1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g
Water was bored. He decided to go through the water cycle. He flew in the air as gas, then condensed into water again, and then rolled down into a river, only to find out that he would be stuck doing it forever
. The end
Answer:
A. Yes, the substance must be water.
Explanation:
The density of a substance is unique to it. Density is defined the as the amount of substance contained per volume.
One of the ways of identifying a substance is to determine its density. Every matter is known to have their own specific densities. This makes them different from other substances. The density of gold is unique to it and it differs from that of silver.
In fact, water has density of 1.00gcm⁻³. Experimental errors and some little factors must have altered our expected figure. This a case of precision and accuracy in the experiment.
Answer:
Inter-molecular forces and molecular volumes are the chief reasons for lower measured pressure
Explanation:
The kinetic theory assumes that gas particles occupy a negligible fraction of the total volume of the gas. It also assumes that the force of attraction between gas molecules is zero.
However, during high pressure, the volume of the gas particles are not negligible compare to the total gas volume and as such the volume of a real gas under such condition is higher than the Ideal gas. Vander-waal attempted to modify the ideal gas equation by subtracting the excess volume from the ideal equation. The increased volume is the reason the measured pressure of a real gas is less than an ideal gas
On the other hand, close to condensation, the other assumption of negligible forces of attraction becomes invalid. As inter-molecular distances decrease, inter-molecular forces increase reducing the bombardment of the wall of the container due to restricted particle movement and lower measured gas pressure.
Answer:
Explanation:
They gave us the masses of two reactants and asked us to determine the mass of the product.
This looks like a limiting reactant problem.
1. Assemble the information
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 239.27 32.00 207.2
2PbS + 3O₂ ⟶ 2Pb + 2SO₃
m/g: 2.54 1.88
2. Calculate the moles of each reactant
3. Calculate the moles of Pb from each reactant
4. Calculate the mass of Pb
